a) \(x^2-4x=0\)

\(x\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

b) \(4x^2-9=0\)

\(\left(2x\right)^2-3^2=0\)

\(\left(2x+3\right)\left(2x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+3=0\\2x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}\)

c) \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\left(x-3\right)\left(2x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

d) \(x\left(2x+9\right)-4x-18=0\)

\(x\left(2x+9\right)-2\left(2x+9\right)=0\)

\(\left(2x+9\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

e) \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)

\(\left(2x-1-x-2\right)\left(2x-1+x+2\right)=0\)

\(\left(x-3\right)\left(3x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}}\)

\(x^2-4x=0\)

\(x.\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\Leftrightarrow x=4\end{cases}}\)

\(4x^2-9=0\)

\(2^2x^2-9=0\)

\(\left(2x\right)^2-9=0\)

\(\left(2x\right)^2-3^2=0\)

\(\Rightarrow\orbr{\begin{cases}\left(2x\right)^2=\left(-3\right)^2\\\left(2x\right)^2=3^2\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}}\)

\(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\left(x-3\right)\cdot\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

\(x\left(2x+9\right)-4x-18=0\)

\(x\left(2x+9\right)-\left(4x+18\right)=0\)

\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)

\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)

\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

\(\left(2x-1\right)^2-\left(x+2\right)^2=0\)

\(\Rightarrow\left(2x-1\right)^2=\left(x+2\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x-1=x+2\\2x-1=-x+2\end{cases}\Rightarrow\orbr{\begin{cases}2x=3+x\\2x=-x+3\end{cases}\Rightarrow\orbr{\begin{cases}2x-x=3\\2x+x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}}}\)

\(\)

\(\text{ 2x+6=0 }\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

\(S=\left\{-3\right\}\)

\(\text{3x-9=0 }\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\)

\(S=\left\{3\right\}\)

\(\text{4x+20=0}\)

\(\Leftrightarrow4x=-20\)

\(\Leftrightarrow x=-5\)

\(S=\left\{-5\right\}\)

\(\text{4x+1=6-x}\)

\(\Leftrightarrow4x+1-6-x=0\)

\(\Leftrightarrow3x-5=0\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

\(S=\left\{\dfrac{5}{3}\right\}\)

a: 2x+6=0

=>2x=-6

=>x=-3

b: 3x-9=0

=>3x=9

=>x=3

c: 4x+20=0

=>x+5=0

=>x=-5

d: 4x+1=6-x

=>5x=5

=>x=1

\(4x+1=6-x\)

\(\Leftrightarrow4x+1-6+x=0\)

\(\Leftrightarrow5x-5=0\)

\(\Leftrightarrow5x=5\)

\(\Leftrightarrow x=1\)

\(S=\left\{1\right\}\)

a: (x^2+9)(9x^2-1)=0

=>9x^2-1=0

=>x^2=1/9

=>x=1/3 hoặc x=-1/3

b: (4x^2-9)(2^(x-1)-1)=0

=>4x^2-9=0 hoặc 2^(x-1)-1=0

=>x^2=9/4 hoặc x-1=0

=>x=1;x=3/2;x=-3/2

c: (3x+2)(9-x^2)=0

=>(3x+2)(3-x)(3+x)=0

=>\(\left[{}\begin{matrix}3x+2=0\\3-x=0\\3+x=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};3;-3\right\}\)

d: (3x+3)^2(4x-4^2)=0

=>3x+3=0 hoặc 4x-16=0

=>x=4 hoặc x=-1

e: \(2^{\left(x-5\right)\left(x+2\right)}=1\)

=>(x-5)(x+2)=0

=>x-5=0 hoặc x+2=0

=>x=5 hoặc x=-2

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

\(\left(x-1\right)3+3x\left(x-1\right)=0\)

<=>  \(3\left(x-1\right)\left(x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy...

a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

a) \(4x^2-1-x\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x-1-x\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

b) \(\left(4x-1\right)^2-9=0\)

\(\Leftrightarrow\left(4x-1-3\right)\left(4x-1+3\right)=0\)

\(\Leftrightarrow\left(4x-4\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-4=0\\4x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{2}\end{cases}}\)

\(\)

a, x=-1/2 hoặc 1

b, x=-1/2 hoặc x=1

P/s sai thì sửa nha

b)=(4x-1)²-3²=(4x-1-3)(4x-1+3)=16x^2-8x-8

c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)

\(\Leftrightarrow2\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=4\)

hay x=5

e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)

\(\Leftrightarrow\left|2x-7\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)

a. ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$

$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$

$\Leftrightarrow x\leq 2$

b. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$

$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$

$\Leftrightarrow 1=2\sqrt{x-2}$

$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$

$\Leftrightarrow \frac{1}{4}=x-2$

$\Leftrightarrow x=\frac{9}{4}$ (tm)

c. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4$

$\Leftrightarrow 2\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}=2$

$\Leftrightarrow x-1=4$

$\Leftrightarrow x=5$ (tm)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4}{9}}\sqrt{x-2}+\sqrt{9}.\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{1}{2}\sqrt{x-2}-\frac{8}{3}\sqrt{x-2}+3\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{5}{6}\sqrt{x-2}-5=0$

$\Leftrightarrow \sqrt{x-2}=6$

$\Leftrightarrow x-2=36$

$\Leftrightarrow x=38$ (tm)

a:

=>\(\sqrt{\left(2x+1\right)^2}=3\)

=>|2x+1|=3

=>2x+1=3 hoặc 2x+1=-3

=>2x=2 hoặc 2x=-4

=>x=-2 hoặc x=1

b: 

ĐKXĐ: x>=1

PT\(\Leftrightarrow5-2\sqrt{x-1}+3\sqrt{x-1}=0\)

=>\(\sqrt{x-1}+5=0\)(vô lý)

Lời giải:

a. PT $\Leftrightarrow \sqrt{(2x+1)^2}=3$

$\Leftrightarrow |2x+1|=3$

$\Leftrightarrow 2x+1=3$ hoặc $2x+1=-3$
$\Leftrightarrow x=1$ hoặc $x=-2$ (tm)

b. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow 5-\sqrt{4(x-1)}+\sqrt{9(x-1)}=0$

$\Leftrightarrow 5-2\sqrt{x-1}+3\sqrt{x-1}=0$

$\Leftrightarrow 5+\sqrt{x-1}=0$
$\Leftrightarrow \sqrt{x-1}=-5<0$ (vô lý)

Do đó không tồn tại $x$ tm.